Consistent hierarchy of axiomatic systems - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:19:45Z http://mathoverflow.net/feeds/question/56180 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56180/consistent-hierarchy-of-axiomatic-systems Consistent hierarchy of axiomatic systems Andreas Thom 2011-02-21T14:54:29Z 2011-02-24T01:47:26Z <p>First of all, I am not an expert in model theory. I just want to get my personal view on the foundations of mathematics straight.</p> <p>I just learned in Sergey Melikhov's <a href="http://mathoverflow.net/questions/34843/what-is-realistic-mathematics/55981#55981" rel="nofollow">answer</a> to another question something about the <a href="http://en.wikipedia.org/wiki/Axiom_of_determinacy" rel="nofollow">Axiom of Determinacy</a> (AD). This Axiom is equivalent to the statement:</p> <p>$\forall G \subseteq Seq(S):$</p> <p>$$\forall a \in S :\exists a' \in S :\forall b \in S :\exists b' \in S :\forall c \in S :\exists c' \in S ... : (a,a',b,b',c,c'...) \in G$$ or $$\exists a \in S :\forall a' \in S :\exists b \in S :\forall b' \in S :\exists c \in S :\forall c' \in S ... :(a,a',b,b',c,c'...) \notin G$$</p> <p>where $Seq(S)$ is the set of all $\omega$-sequences of some countable set $S$. It is thus some infinitary generalization of de Morgan's rule and seems rather natural. $ZF+AD$ seems to be rather <em>realistic</em> (in my opinion) version of set theory, in the sense that it avoids many <em>unphysical</em> paradoxes (such as non-measurable sets, paradoxical decompositions, non-continuous linear functions on Banach spaces etc.) Still, it seems to be strong enough to reproduce enough infinitary mathematics, so that a development a lot of mathematics and of theoretical physics etc. is possible.</p> <p>It is a fact that $ZF+AD$ can prove consistency of $ZF$. Basically, the question is whether one can continue this process. The question is more precisely:</p> <blockquote> <p><strong>Question:</strong> Is there a hierarchy, which consists of an axiom $AD_{\alpha}$ for <strike> each ordinal</strike> <strong>some ordinals</strong> $\alpha$ (inspired maybe by some version of de Morgan's laws for ordinal sequences of quantifiers applied to sets of sequences in sets of some cardinality), such that $AD=AD_{1}$ and $ZF + \cup_{\beta \leq \alpha} ZF_{\beta}$ can prove consistency of $ZF + \cup_{\beta &lt; \alpha} ZF_{\beta}$.</p> </blockquote> <p>If that is the case, why not taking $ZF + \cup_{\alpha} AD_{\alpha}$ as the axiomatic foundation of mathematics. <strike> The bad thing about this would be that the axioms do not form a set, the payoff would be that it proves the consistency of itself and is philosophically sound.</strike></p> <p><strong>EDIT:</strong> From Emil JeÅ™ábek's comment I understand that the question does not make sense as stated since there are only countably many formulas and hence there cannot be uncountably many axioms. So the right (and more modest) question to ask would maybe be the following:</p> <blockquote> <p><strong>Question:</strong> Does there exist an equally natural axiom $AD'$ (based again on some infinitary version of a well-known principle like de Morgan's laws) which proves (together with $ZF+AD$) consistency $ZF + AD$?</p> </blockquote> <p>Maybe there is also some infinitary version of set theory which allows for sentences of arbitrary length like the one which was used above to describe the meaning of $AD$. This could be the place, where the first question could have an answer.</p> http://mathoverflow.net/questions/56180/consistent-hierarchy-of-axiomatic-systems/56198#56198 Answer by David FernandezBreton for Consistent hierarchy of axiomatic systems David FernandezBreton 2011-02-21T18:59:50Z 2011-02-21T20:07:32Z <p>Regarding the sentences of arbitrary length, they can be expressed with the help of <a href="http://en.wikipedia.org/wiki/Infinitary_logic" rel="nofollow">infinitary logics</a>. Roughly, the language $L_{\kappa,\lambda}$ is similar to other formal languages except for the fact that it allows any conjunction or disjunction of less than $\kappa$ sentences (and provides for an infinitary form of De Morgan's rules), and any chain of less than $\lambda$ quantifiers. So your informal $\forall a\in S:\exists a'\in S\cdots$ can be expressed properly in, say, $L_{\omega_1,\omega}$. However, I have no idea on whether someone has considered any axiom system analogous to ZFC which uses this language...</p> http://mathoverflow.net/questions/56180/consistent-hierarchy-of-axiomatic-systems/56475#56475 Answer by Jason for Consistent hierarchy of axiomatic systems Jason 2011-02-24T01:47:26Z 2011-02-24T01:47:26Z <p>There is a generalization of the compactness theorem to infinitary logics that sounds somewhat close to what you want. Specifically, the compactness theorem tells us that every finitely satisfiable theory of the usual formulas from a first-order language $L$ is satisfiable. For any uncountable cardinal $\kappa$, we can extend $L = L_{\omega, \omega}$ to the infinitary language $L_{\kappa, \kappa}$ by closing the usual formulas under $\bigwedge_{\xi &lt; \lambda}\varphi_{\xi}$, $\bigvee_{\xi &lt; \lambda}\varphi_{\xi}$, $\exists \langle x_{\xi}| \xi &lt; \lambda\rangle$, and $\forall \langle x_{\xi}| \xi &lt; \lambda\rangle$ for all $\lambda &lt; \kappa$. We then define $\kappa$ to be strongly compact if an analogous theorem holds for arbitrary $L_{\kappa, \kappa}$, mainly that if every collection of fewer than $\kappa$ many statements from a theory in $L_{\kappa, \kappa}$ is satisfiable, then the theory is satisfiable.</p> <p>Now let me emphasize that while ZF + AD may sound natural, it does much much more than prove the consistency of ZF. In fact, ZF + AD proves the consistency of ZFC + "There exists a Woodin cardinal", and Woodin cardinals are quite high up in the large cardinal hierarchy (i.e., a sufficiently stronger theory than ZF that is more likely to be inconsistent than ZF alone). Also, since set theorists tend to like having choice, the usual assumption is not that AD holds in our universe. Instead, we assume that it holds in the minimal transitive ZF model containing all of the ordinals and all of the reals, i.e., $L(\mathbb{R})$.</p> <p>Now from ZFC + "There exists a strongly compact cardinal", we can prove that ZF + AD holds in $L(\mathbb{R})$. Moreover, we can prove that for every $\alpha$, there exists $\beta > \alpha$ such that AD holds in the set $L_{\beta}(\mathbb{R})$. In particular, ZFC + "There exists a strongly compact cardinal" proves CON(ZF + AD), CON(ZF + AD + CON(ZF + AD)), CON(ZF + AD + CON(ZF + AD + CON(ZF + AD))), etc. </p> <p>The above result illustrates how assuming a strongly compact cardinal is also a very strong large cardinal hypothesis, but it is only one (admittedly significant) jump higher in our large cardinal hierarchy.</p>